Your mistake is in the 3rd to last step, where you have one term with a 2x-3y^2 denominator. In the following step you simplified the equation by multiplying (2x-3y^2) with the (2x-3y^2)^2 term. However the two 4y terms on the numerator were not being divided by 2x-3y^2.

Okay, I was able to get the answer in the right form as the provided answer. However, it is off numerically by 2. The provided answer has 18xy and my answer has 16xy. I have always known 8 + 8 to equal 16...what gives?

Staff: Mentor

Some comments.
1) You don't take "f' " of an expression. f'(x) is the derivative of f. The operator is d/dx. f' and f'' should not appear in your work at all.
2) You wrote f(x) = 2xy - y3 = 0. This is incorrect because a) the expression 2xy - y3 involves both x and y, not just x alone, and b) the equation 2xy - y3 = 0 defines an implicit relationship between x and y.

since it is a function of two variables? Then we would just need to find [itex]\frac{\partial f}{\partial x}[/itex] and [itex]\frac{\partial f}{\partial y}[/itex]. If this isn't the case, we would just have

[itex]2xy - y^{3} = 5[/itex],

and we would need to find [itex]\frac{dy}{dx}[/itex] implicitly, not f'(x).

since it is a function of two variables? Then we would just need to find [itex]\frac{\partial f}{\partial x}[/itex] and [itex]\frac{\partial f}{\partial y}[/itex]. If this isn't the case, we would just have

[itex]2xy - y^{3} = 5[/itex],

and we would need to find [itex]\frac{dy}{dx}[/itex] implicitly, not f'(x).

So guys, which is it? Or am I completely off on this one?

No, I think you are on track here. f'(x) is meaningless here, since the left side of the equation 2xy - y3 - 5 = 0 represents a function of two variables.

The tacit assumption here, I believe, is that y is an implicit function of x, and goal is to find dy/dx using implicit differentiation. Note that this should be written as dy/dx = ..., not f'(x) = ..., so as to not cause confusion between the function (of two variables) in the first equation, and the function (of one variable) that relates x and y.